Maclaurin series for $f(x)=\frac {1}{(1+x+x^2)} $ Find the Maclaurin series of $f(x)=\dfrac {1}{(1+x+x^2)} $and the radius of convergence of the series.
$f(x) = \dfrac {(1-x)}{(1-x^3)}
 = (1-x) \sum _{0} ^\infty x^{(3n)}$
and so?
How this can be the shape of $\sum_0 ^\infty 
\dfrac {f^n}{n!}x^n$    ?
And I don't know how to get a radius.
 A: Ok, taking it from where you leave it:
$$f(x)=(1-x)\sum_{n=0}^\infty x^{3n}=\sum_{n=0}^\infty x^{3n}(1-x)=$$
For the radius of convergence - quotient test:
$$\left|\frac{\;x^{3n+3}(1-x)}{x^{3n}(1-x)}\;\right|=|x|^3<1\implies |x|<1\ldots$$
A: Calculate 
$$(1-x)(1+x^3+x^6+x^9+x^{12}+\cdots).$$
We get
$$1-x+x^3-x^4+x^6-x^7+x^9-x^{10}+x^{12}-x^{13}+\cdots.$$
For the radius of convergence, note that the terms of the series do not approach $0$ if $|x|\ge 1$. So the series diverges if $|x|\ge 1$.
We will prove (absolute) convergence if $|x|\lt 1$.
Take the absolute values of the terms. This gives us the series
$$1+|x|+|x|^3+|x|^4+|x|^6+|x|^7+|x|^9+\cdots.\tag{1}$$
By comparison with the geometric series
$$1+|x|+|x|^2 +|x|^3+|x|^4+|x|^5+|x|^6+\cdots,$$
which converges if $|x|\lt 1$, we can see that (1) converges if $|x|\lt 1$. It follows that the radius of convergence of our series is $1$. 
Remark: Let $f(x)=\frac{1}{1+x+x^2}$. The series we obtained does indeed have the shape $\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$. However, computing all but the first few derivatives of $f$ is fairly painful, which is one reason that the series for $\frac{1}{1+x+x^2}$ was obtained in a somewhat indirect way. 
